Problem: $f(x)=(x-6)^2-12.25$ 1) What are the zeros of the function? Write the smaller $x$ first, and the larger $x$ second. $\text{smaller }x=$
$\begin{aligned} (x-6)^2-12.25&=0 \\\\ (x-6)^2&=12.25 \\\\ \sqrt{(x-6)^2}&=\sqrt{12.25} \\\\ x-6&=\pm 3.5 \\\\ x&=\pm3.5+6 \\\\ x={2.5}&\text{ or }x={9.5} \end{aligned}$ $f(x)$ is given in vertex form: $f(x)=(x-{6})^2{-12.25}$ So the vertex of the parabola is at $({6},{-12.25})$. In conclusion, $\begin{aligned} \text{smaller }x&=2.5 \\\\ \text{larger }x&=9.5 \end{aligned}$ The vertex of the parabola is at $(6,-12.25)$